Problem 101 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=\left\\{\begin{array}{cc}\int_{0}^{x}(5+|1-t|) d t, x>2 \\ 5 x+1, x \leq 2\end{array}\right.\), then at \(x=2\) (A) \(f(x)\) is continuous (B) \(f(x)\) is not continuous (C) \(f(x)\) is differentiable (D) \(f(x)\) is not differentiable

Step-by-Step Solution

Answer

(A) and (D) are correct; \(f(x)\) is continuous but not differentiable at \(x=2\).

Step 1: Understand the Function

The function \(f(x)\) is defined piecewise with two expressions; one involving an integral for \(x > 2\) and the other a linear expression for \(x \leq 2\). We need to analyze both parts to check the continuity and differentiability at \(x = 2\).

Step 2: Analyze Continuity

To check continuity at \(x = 2\), calculate whether \(\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) = f(2)\). Given \(f(x) = 5x + 1\) for \(x \leq 2\), we have \(f(2) = 5 \times 2 + 1 = 11\). We must evaluate the integral part from the right.

Step 3: Compute Integral for Right-hand Limit

Calculate \( \lim_{{x \to 2^+}} \int_{0}^{x} (5 + |1-t|) \, dt \). Divide the integral into parts based on \(|1-t|\), as it behaves differently depending on \(t\).

Step 4: Evaluate Piecewise Integral

For \(t > 1\), \(|1-t| = t-1\). So, split the integral \(\int_{0}^{2} (5 + |1-t|) \, dt = \int_{0}^{1} (5 + 1-t) \, dt + \int_{1}^{2} (5 + t-1) \, dt\).

Step 5: Solve the First Integral

Evaluate \(I_1 = \int_{0}^{1} (6-t) \, dt = [6t - \frac{t^2}{2}]_{0}^{1} = 6 \times 1 - \frac{1^2}{2} = 6 - 0.5 = 5.5\).

Step 6: Solve the Second Integral

Evaluate \(I_2 = \int_{1}^{2} (4 + t) \, dt = [4t + \frac{t^2}{2}]_{1}^{2} = (8 + 2) - (4 + 0.5) = 10 - 4.5 = 5.5\).

Step 7: Combine Integrals

Combine the two parts: \(I_1 + I_2 = 5.5 + 5.5 = 11\). Thus, \(\lim_{{x \to 2^+}} f(x) = 11\).

Step 8: Check Continuity

Compare the left and right limits to \(f(2): \lim_{{x \to 2^-}} f(x) = 11, \lim_{{x \to 2^+}} f(x) = 11, f(2) = 11\). Therefore, \(f(x)\) is continuous at \(x = 2\).

Step 9: Analyze Differentiability

For differentiability, we require the left and right derivatives to be equal at \(x = 2\). Compute \(f'(x)\) for \(x \leq 2\) and \(x > 2\) and check the limits.

Step 10: Compute Left Derivative

For \(x \leq 2\), \(f'(x) = \frac{d}{dx}(5x+1) = 5\).

Step 11: Compute Right Derivative

Compute \(\frac{d}{dx}(\int_{0}^{x} (5 + |1-t|) \, dt) = 5 + |1-x|\). For \(x = 2\), \(|1-2| = 1\), so \(f'(2^+) = 5 + 1 = 6\).

Step 12: Check Differentiability

Compare the left and right derivatives: \(f'(2^-) = 5\) and \(f'(2^+) = 6\). Since these are not equal, \(f(x)\) is not differentiable at \(x = 2\).

Key Concepts

Continuity

Continuity is a fundamental property of functions indicating that a small change in the input results in a small change in the output. For a function to be continuous at a point, the following must hold: the left-hand limit, right-hand limit, and the actual value of the function at that point should all be equal.
In our piecewise function problem, we are tasked with evaluating the continuity of the function \( f(x) \) at \( x = 2 \). The function is defined differently on either side of \( x = 2 \), necessitating a close examination of both parts.
To check the continuity at \( x = 2 \), we need to calculate both the left-hand limit \( \lim_{{x \to 2^-}} f(x) \) and the right-hand limit \( \lim_{{x \to 2^+}} f(x) \), and compare these with the function's value at \( x = 2 \), which is \( f(2) = 11 \).
Upon evaluating, both limits equate to 11, thus confirming that \( f(x) \) is indeed continuous at the point \( x = 2 \). It is important to understand that continuity ensures the graph of the function has no breaks, jumps, or holes at the specified point.

Differentiability

Differentiability is concerned with the ability to find the derivative, or the instantaneous rate of change, at any given point on a function. For a function to be differentiable at a point, the left-hand derivative and right-hand derivative at that point must be equal. An important thing to remember is that differentiability implies continuity, but the reverse is not always true.
In the case of our piecewise function at \( x = 2 \), we examined the derivative for each segment of the function. The left derivative, calculated from \( f(x) = 5x + 1 \), results in \( f'(2^-) = 5 \). Meanwhile, for the right segment involving the integral, our right derivative equals \( f'(2^+) = 6 \).
Since these derivatives are not equal, \( f(x) \) is not differentiable at \( x = 2 \). Remember, a cusp or a sharp turn in the graph of \( f(x) \) at a given point indicates non-differentiability.

Integrals

Integrals are fundamental in mathematics for finding the accumulation of quantities, such as areas under curves or total changes. In our exercise, the integral defines \( f(x) \) for values greater than 2 and provides a way to smooth out and calculate the accumulation of change described by the function.
The integral for the right side of \( f(x) \) is expressed as \( \int_{0}^{x} (5 + |1-t|) \, dt \). Calculating it involves breaking it down into manageable sections based on the behavior of \( |1-t| \). This gives rise to two separate integrals from which their results need to be combined to complete the evaluation.
The process of solving these integral segments involved evaluating:

• \( \int_{0}^{1} (6-t) \, dt \)
• \( \int_{1}^{2} (4+t) \, dt \)

Both integrals result in a sum of 11, which matches the left-hand value, leading us to confidently establish the limits of continuity.

Limits

Limits are fundamental to understanding behavior in calculus, especially when analyzing continuity and differentiability. They describe what happens to a function as the input approaches a particular value.
For our piecewise function, limits play a crucial role in evaluating the behavior of \( f(x) \) at \( x = 2 \). Only if both the left-hand limit \( \lim_{{x \to 2^-}} f(x) \) and the right-hand limit \( \lim_{{x \to 2^+}} f(x) \) meet and equal \( f(2) \), can \( f(x) \) be continuous.
In our analysis, calculating the limit from the integral side involves assessing the piecewise nature of the function and solving stepwise integrals, ensuring consistency. Both sides meet the value of 11, indicating continuity at this point.
Understanding limits allows us to predict the function's behavior at critical points, ensuring thorough comprehension of its entire structure around specific inputs.